If x satisfies the equation log2(1+x4)+log2(1−x+44)=2log2(x−12−1), then find the sum of squares of values of x.

log2(xx+4)+log2(x+4x)=2log2(x−13−x)⇒2log2(xx+4)=2log2(x−13−x)⇒log(xx+4)=±log(x−13−x)Case 1: ⇒xx+4=x−13−x⇒x=±2⇒x=2 is the only solution since it lies the domain. Case 2:xx+4=3−xx−1⇒x=±6⇒x=6, lies in the domain. Required Answer =22+62=8

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If x satisfies the equation ^{2} ( 1+{4}{x} )+ ^{2} ( 1-{4}

Question

If $x$ satisfies the equation $lo g^{2} (1 + \frac{4}{x}) + lo g^{2} (1 - \frac{4}{x + 4}) = 2 lo g^{2} (\frac{2}{x - 1} - 1)$ , then find the sum of squares of values of $x$ .

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x

satisfies the equation

lo g^{2} (1 + \frac{4}{x}) + lo g^{2} (1 - \frac{4}{x + 4}) = 2 lo g^{2} (\frac{2}{x - 1} - 1)

, then find the sum of squares of values of

x

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Text solutionVerified

lo g^{2} (\frac{x + 4}{x}) + lo g^{2} (\frac{x}{x + 4}) = 2 lo g^{2} (\frac{3 - x}{x - 1})

\Rightarrow 2 lo g^{2} (\frac{x + 4}{x}) = 2 lo g^{2} (\frac{3 - x}{x - 1})

\Rightarrow lo g (\frac{x + 4}{x}) = \pm lo g (\frac{3 - x}{x - 1})

Case 1:

\Rightarrow \frac{x + 4}{x} = \frac{3 - x}{x - 1}

\Rightarrow x = \pm 2

\Rightarrow x = 2

is the only solution since it lies the domain.
Case 2:

\frac{x + 4}{x} = \frac{x - 1}{3 - x}

\Rightarrow x = \pm 6

\Rightarrow x = 6

, lies in the domain.

Required Answer

= 2^{2} + 6^{2} = 8

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Question Text	If $x$ satisfies the equation $lo g^{2} (1 + \frac{4}{x}) + lo g^{2} (1 - \frac{4}{x + 4}) = 2 lo g^{2} (\frac{2}{x - 1} - 1)$ , then find the sum of squares of values of $x$ .
Answer Type	Text solution:1
Upvotes	150

If x satisfies the equation log2(1+x4​)+log2(1−x+44​)=2log2(x−12​−1), then find the sum of squares of values of x.

Text solutionVerified

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If $x$ satisfies the equation $lo g^{2} (1 + \frac{4}{x}) + lo g^{2} (1 - \frac{4}{x + 4}) = 2 lo g^{2} (\frac{2}{x - 1} - 1)$ , then find the sum of squares of values of $x$ .